A question on volume I found the following problem (#2 of SIMOC 2015 Sample Papers – Grade 10(Secondary 4)).

The answer key is $V \gt 18 \pi$.
In order to get the said answer, I can use brute force by setting up
$$\dfrac {V}{(4/6)\pi 10^3} \gt (\dfrac {3}{10})^3$$
1) Can anyone tell me why the above (using ratio of volumes of “similar” objects) work? 
2) Would it be a bit unfair to the Secondary 4 students because the relation could have been coming from either $V \lt \pi (10^2 – 7^2) \times 3$ or $\pi (10^2 – 7^2) \times 7 \lt (4/6)\pi 10^3 – V$ or some other comparisons with the respective cylindrical objects. Maybe eventually the students have found that none of the comparisons meet the given options, but time has been wasted. Afterall, this is just a MC question.
 A: Consider hemisphere of radius $3$. The volume of such a hemisphere is $\frac{1}{2}\frac{4}{3}(3)^3\pi = 18\pi$. This hemisphere has volume smaller than the volume of the water, and so $V>18\pi$.
It doesn't sound outlandish to me to pose this question to fourth graders, especially fourth graders doing competition mathematics. The problem is computationally straightforward, requires almost no reasoning, and the only requires knowing the volume of a sphere. I fully expect clever 10-year olds to be able to solve this, especially with practice.

I think your equation is supposed to read: $$\frac{\text{volume of smaller}}{\text{volume of larger}}=\dfrac {V}{(4/3)\pi 10^3} \gt (\dfrac {3}{10})^3 = \text{(scale factor)}^3$$
Notice that the water is not similar to the whole shape, because the water is not hemispherical. But, as my explanation shows, $V$ is greater than the volume of the hemisphere of radius $3$ that would be similar to the hemisphere. When converting between the volumes of similar shapes, you need to multiply by the scale factor cubed, so if $V$ was the volume of the similar hemisphere then the above inequality would be an equality. Since $V$ is in fact larger, it is in inequality with the left hand side being larger. That's why the inequality holds. Solving for $V$ gives $V>18\pi$ as desired. I’m not sure why you describe this as “brute force.”
A: A hint: 
Using one of the standard methods to compute the volume of rotational bodies one obtains
$$V=\pi\int_7^{10}(100-x^2)\>dx\ .$$
